Question
Question: What is the derivative of \(\tanh \left( x \right)\)?...
What is the derivative of tanh(x)?
Solution
In this question we have to find the derivative of the hyperbolic function. In mathematics, the hyperbolic functions are similar to the trigonometric functions or circular functions. The hyperbolic functions are analogs of the circular function or the trigonometric functions. The basic hyperbolic functions are hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh).
Complete step by step solution:
Hyperbolic functions work in the same way as the “normal” trigonometric “cuisine” but instead of referring to a unit circle, they refer to a set of hyperbolas. If we compare the derivatives of hyperbolic functions with the derivatives of the standard trigonometric functions, then we find lots of similarities and differences also. For example, the derivatives of the sine functions match:
⇒dxdsinx=cosx⇒dxdsinhx=coshx
But we can see difference also
⇒dxdcosx=−sinx⇒dxdcoshx=sinhx
The hyperbolic functions are defined through the algebraic expression that includes the exponential function ex and its inverse exponential functione−x. To find the derivative of the tanhx, we will use the trigonometric rule which is coshxsinhx=tanhx.
Now the value of sinhx in term of exponential function is:
⇒sinhx=21(ex−e−x)
Similarly, the value of coshx in term of exponential function is:
⇒coshx=21(ex+e−x)
Now,
⇒tanhx=coshxsinhx⇒tanhx=21(ex+e−x)21(ex−e−x)⇒tanhx=(ex+e−x)(ex−e−x)
Now we will differentiate it with respect to x. since we can easily see that the above expression will be differentiated by quotient rule.
We know the quotient rule is:
⇒dxd(vu)=v2dxduv−dxdvu
We also know the derivative of ex is ex, and derivative of e−x is −e−x .
Now applying the quotient rule, we get
⇒dxd(tanhx)=(ex+e−x)2(ex−e−x) ˋ(ex+e−x)−(ex−e−x)(ex+e−x) ˋ⇒dxd(tanhx)=(ex+e−x)2(ex+e−x)(ex+e−x)−(ex−e−x)(ex−e−x)
Now by more simplifying, we get